RD Sharma 2020 solution class 9 chapter 24 Measures of Central Tendency MCQS

MCQS

Page-24.20

Question 1:

Mark the correct alternative in each of the following:
Which one of the following is not a measure of central value?

(a) Mean

(b) Range

(c) Median

(d) Mode

Answer 1:

We know that mean, median and mode are all measures of central tendency.

Hence, the correct choice is (b).

Question 2:

The mean of n observations is X. If k is added to each observation, then the new mean is

(a) X

(b) X + k

(c) X − k

(d) kX

Answer 2:

Let us take n observations.

Ifbe the mean of the n observations, then we have

Add a constant k to each of the observations. Then the observations becomes

Ifbe the mean of the new observations, then we have

Hence, the correct choice is (b).

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Question 3:

The mean of n observations is X. If each observation is multiplied by k, the mean of new observations is

(a) $k\overline{X}$

(b) $\frac{\overline{X}}{k}$

(c) $\overline{X}+k$

(d) $\overline{X}-k$

Answer 3:

Let us take n observations.

Ifbe the mean of the n observations, then we have

Multiply a constant k to each of the observations. Then the observations becomes

Ifbe the mean of the new observations, then we have

Hence, the correct choice is (a).

Question 4:

The mean of a set of seven numbers is 81. If one of the numbers is discarded, the mean of the remaining numbers is 78. The value of discarded number is

(a) 98

(b) 99

(c) 100

(d) 101
 

Answer 4:

Given that the mean of 7 numbers is 81. Let us denote the numbers by.

Ifbe the mean of the n observations, then we have

Hence the sum of 7 numbers is

If one number is discarded then the mean becomes 78 and the total numbers becomes 6.

Let the number discarded is x.

After discarding one number the sum becomesand then the mean is

But it is given that after discarding one number the mean becomes 78.

Hence we have

Thus the excluded number is. So, the correct choice is (b).

Question 5:

For which set of numbers do the mean, median and mode all have the same value?
(a) 2, 2, 2, 2, 4

(b) 1, 3, 3, 3, 5

(c) 1, 1, 2, 5, 6

(d) 1, 1, 1, 2, 5

Answer 5:

For the data 2, 2, 2, 2, 4 of 5 numbers, we have

Since, 2 occurs maximum number of times,

For the data 1, 3, 3, 3, 5 of 5 numbers, we have

Since, 3 occurs maximum number of times,

Hence, the correct choice is (b).

Note that if it happens that the result is not true for the second data then we must follow the same procedure for the other data’s.

Question 6:

For the set of numbers 2, 2, 4, 5 and 12, which of the following statements is true?

(a) Mean = Median

(b) Mean > Mode

(c) Mean > Mode

(d) Mode = Median

Answer 6:

For the data 2, 2, 4, 5, 12 of 5 numbers, we have

Since, 2 occurs maximum number of times,

So, the correct choice is (b).

Question 7:

If the arithmetic mean of 7, 5, 13, x and 9 is 10, then the value x is

(a) 10

(b) 12

(c) 14

(d) 16

Answer 7:

The given data is 7, 5, 13, x and 9. They are 5 in numbers.

The mean is

But, it is given that the mean is 10. Hence, we have

Hence, the correct choice is (d).

Question 8:

If the mean of five observations x, x+2, x+4, x+6, x+8, is 11, then the mean of first three observations is

(a) 9

(b) 11

(c) 13

(d) none of these

Answer 8:

The given data is x, x + 2,  x + 4, x + 6 and x + 8. They are 5 in numbers.

The mean is

But, it is given that the mean is 11. Hence, we have

Then the first three observations are 7, 7 + 2, 7+4, that is, 7, 9, 11. Their mean is

Hence, the correct choice is (a).

Question 9:

Mode is

(a) least frequent value

(b) middle most value

(c) most frequent value

(d) none of these

Answer 9:

We know that, mode is the observation which occur maximum number of times.

Hence, the correct choice is (c).

Question 10:

The following is the data of wages per day : 5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10, 8
The mode of the data is

(a) 7

(b) 5

(c) 8

(d) 10

Answer 10:

The given data is 5, 4, 7, 5, 8, 8, 8, 5, 7, 9, 5, 7, 9, 10 and 8.

Make the following frequency table.

Since the values 5 and 8 occurs in the data maximum number of times, that is, 4. Hence, the modal value is 5 and 8. In this case the mode is not unique.

Hence, the correct options are (b) and (c).

Question 11:

The median of the following data : 0, 2, 2, 2, −3, 5, −1, 5, −3, 6, 6, 5, 6 is

(a) 0

(b) −1.5

(c) 2

(d) 3.5

Answer 11:

The given data is 0, 2, 2, 2, -3, 5, -1, 5, 5, -3, 6, 6, 5 and 6.

Arranging the given data in ascending order, we have

-3, -3, -1, 0, 2, 2, 2, 5, 5, 5, 5, 6, 6, 6

Here, the number of observation, which is an even number.

Hence, the median is

So, the correct choice is (d).

Question 12:

The algebraic sum of the deviations of a set of n values from their mean is

(a) 0

(b) n − 1

(c) n

(d) n + 1

Answer 12:

Ifbe the mean of the n observations, then we have

Letbe the mean of n values. So, we have

The sum of the deviations of n valuesfrom their meanis

Hence the correct choice is (a).

Question 13:

A, B, C are three sets of values of x:

(a) A: 2, 3, 7, 1, 3, 2, 3
(b) 7, 5, 9, 12, 5, 3, 8
(c) 4, 4, 11, 7, 2, 3, 4

Which one of the following statements is correct?

(a) Mean of A = Mode of C

(b) Mean of C = Median of B

(c) Median of B = Mode of A

(d)Mean, Median and Mode of A are equal.
 

Answer 13:

For the data A: 2, 3, 7, 1, 3, 2, 3 of 7 numbers, we have

Arranging the data A in ascending order, we have

A: 1, 2, 2, 3, 3, 3, 7

Since, 3 occurs maximum number of times,

Hence, the correct choice is (d).

Note that if it happens that the result is not found in the first step then we must follow the same procedure for the other data’s.

Question 14:

The mean of five number is 30. If one number is excluded, their mean becomes 28. The excluded number is 

(a) 28
(b) 30
(c) 35
(d) 38

Answer 14:

We know

Mean = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

⇒ Sum of observations = Mean × Number of observations

Mean of five numbers = 30     (Given)

∴ Sum of five numbers = 30 × 5 = 150        .....(1)

If one number is excluded, their mean is 28.

∴ Mean of four numbers = 28

⇒ Sum of four numbers = 28 × 4 = 112        .....(2)

Now,

Excluded number

= Sum of five numbers − Sum of four numbers

= 150 − 112                        [From (1) and (2)]

= 38

Thus, the excluded number is 38.

Hence, the correct answer is option (d).

Question 15:

If the mean of the observation : x, x + 3, x + 5, x + 7, x + 10 is 9, the mean of last three observation is 

$$\begin{gathered} \left(\mathrm{a}\right) \frac{31}{3} \\ \left(\mathrm{b}\right) \frac{32}{3} \\ \left(\mathrm{c}\right) \frac{34}{3} \\ \left(\mathrm{d}\right) \frac{35}{3} \end{gathered}$$

Answer 15:

The given observations are x, x + 3, x + 5, x + 7, x + 10.

Mean of the given observations = 9

Number of observations = 5

Now,

$\mathrm{Mean}=\frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

$\Rightarrow 9=\frac{x+\left(x+3\right)+\left(x+5\right)+\left(x+7\right)+\left(x+10\right)}{5}$

$\Rightarrow 5x+25=45$

$\Rightarrow 5x=45-25=20$

$\Rightarrow x=\frac{20}{5}=4$

∴ x + 5 = 4 + 5 = 9

x + 7 = 4 + 7 = 11

x + 10 = 4 + 10 = 14

So, the last three observations are 9, 11 and 14.

∴ Mean of the last three observations = $\frac{9+11+14}{3}=\frac{34}{3}$                        $\left(\mathrm{Mean}=\frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}\right)$

Thus, the mean of the last three observations is $\frac{34}{3}$.

Hence, the correct answer is option (c).

Question 16:

If $\overline{X}$ is the mean of x₁,x₂,.......,x_n , then for $a\ne 0$, the mean of  ax₁,ax₂,......,ax_n,  $\frac{x_1}{a},\frac{x_2}{a},...,\frac{x_n}{n} \mathrm{is}$

$$\begin{gathered} \left(\mathrm{a}\right) \left(a+\frac{1}{a}\right)\overline{\mathrm{X}} \\ \left(\mathrm{b}\right) \frac{1}{2}\left(a+\frac{1}{a}\right)\overline{\mathrm{X}} \\ \left(\mathrm{c}\right) \left(a+\frac{1}{n}\right)\frac{\overline{\mathrm{X}}}{n} \\ \left(\mathrm{d}\right) \left(a+\frac{1}{a}\right)\frac{\overline{\mathrm{X}}}{2n} \end{gathered}$$

Answer 16:

It is given that, the mean of x₁, x₂,..., x_n is $\overline{X}$.

$\therefore \overline{X}=\frac{x_1+x_2+...+x_n}{n}$                   $\left(\mathrm{Mean}=\frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}\right)$
   
$\Rightarrow x_1+x_2+...+x_n=n\overline{X}$        .....(1)

Now, $a{x}_1,a{x}_2,...,a{x}_n,\frac{x_1}{a},\frac{x_2}{a},...,\frac{x_n}{a}$ are 2n observations.

∴ Mean of these observations

$=\frac{a{x}_1+a{x}_2+...+a{x}_n+{\displaystyle \frac{x_1}{a}}+{\displaystyle \frac{x_2}{a}}+...+{\displaystyle \frac{x_n}{a}}}{2n}$

$=\frac{a\left(x_1+x_2+...+x_n\right)+{\displaystyle \frac{1}{a}}\left(x_1+x_2+...+x_n\right)}{2n}$

$=\frac{\left(a+{\displaystyle \frac{1}{a}}\right)\left(x_1+x_2+...+x_n\right)}{2n}$

$=\frac{\left(a+{\displaystyle \frac{1}{a}}\right)\times n\overline{X}}{2n}$                [Using (1)]

$=\frac{1}{2}\left(a+\frac{1}{a}\right)\overline{X}$

Thus, the mean of $a{x}_1,a{x}_2,...,a{x}_n,\frac{x_1}{a},\frac{x_2}{a},...,\frac{x_n}{a}$ is $\frac{1}{2}\left(a+\frac{1}{a}\right)\overline{X}$.

Hence, the correct answer is option (b).

Question 17:

Let $\overline{\mathrm{X}}$ be the mean of x₁,x₂,...,x_n, and $\overline{\mathrm{Y}}$ the mean of  y₁,y₂,...,y_n, if  $\overline{\mathrm{Z}}$ is the mean of  x₁,x₂,...,x_n, y₁,y₂,...,y_n, then $\overline{\mathrm{Z}}$ is equal to 

$$\begin{gathered} \left(a\right) \overline{\mathrm{X}} +\overline{\mathrm{Y}} \\ \left(b\right) \frac{\overline{\mathrm{X}}+\overline{\mathrm{Y}}}{2} \\ \left(\mathrm{c}\right) \frac{\overline{\mathrm{X}}+\overline{\mathrm{Y}}}{n} \\ \left(\mathrm{d}\right) \frac{\overline{\mathrm{X}}+\overline{\mathrm{Y}}}{2n} \end{gathered}$$

Answer 17:

It is given that, the mean of x₁, x₂,..., x_n is $\overline{X}$.

$\therefore \overline{X}=\frac{x_1+x_2+...+x_n}{n}$           $\left(\mathrm{Mean}=\frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}\right)$
   
$\Rightarrow x_1+x_2+...+x_n=n\overline{X}$        .....(1)

Also, the mean of y₁, y₂,..., y_n is $\overline{Y}$.

$\therefore \overline{Y}=\frac{y_1+y_2+...+y_n}{n}$

$\Rightarrow y_1+y_2+...+y_n=n\overline{Y}$        .....(2)

Now, $x_1,x_2,...,x_n,y_1,y_2,...,y_n$ are 2n observations. The mean of these 2n observations is $\overline{Z}$.

$\therefore \overline{Z}=\frac{x_1+x_2+...+x_n+y_1+y_2+...+y_n}{2n}$

$\Rightarrow \overline{Z}=\frac{n\overline{X}+n\overline{Y}}{2n}$                [Using (1) and (2)]

$\Rightarrow \overline{Z}=\frac{n\left(\overline{X}+\overline{Y}\right)}{2n}$

$\Rightarrow \overline{Z}=\frac{\overline{X}+\overline{Y}}{2}$

Hence, the correct answer is option (b).

Question 18:

If ${\overline{\mathrm{X}}}_1,{\overline{\mathrm{X}}}_2,...,{\overline{\mathrm{X}}}_k$ are the means of  n group with n1,n2,...,nk number  of observations respectively, then the mean $\overline{\mathrm{X}}$ of all the groups taken together is govern by 

$$\begin{gathered} \left(\mathrm{a}\right) \sum _{i=1}^k{n}_i {\overline{\mathrm{X}}}_i \\ \left(\mathrm{b}\right) \frac{1}{n^2}\sum _{i=1}^k n_i {\overline{\mathrm{X}}}_i \\ \left(\mathrm{c}\right) \frac{{\displaystyle \sum _{i=1}^k} n_i {\overline{\mathrm{X}}}_i}{{\displaystyle \sum _{i=1}^k} n_i} \\ \left(\mathrm{d}\right) \frac{{\displaystyle \sum _{i=1}^k n_i {\overline{\mathrm{X}}}_i}}{{\displaystyle 2_n}} \end{gathered}$$

Answer 18:

We know

Mean = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

⇒ Sum of observations = Mean × Number of observations


Mean of n₁ observations of first group = $\overline{X_1}$

∴ Sum of n₁ observations of first group = $n_1\overline{X_1}$

Mean of n₂ observations of second group = $\overline{X_2}$

∴ Sum of n₂ observations of second group = $n_2\overline{X_2}$

.            .            .            .            .            .            .
.            .            .            .            .            .            .

Mean of n_k observations of kth group = $\overline{X_k}$

∴ Sum of n_k observations of kth group = $n_k\overline{X_k}$

Now,

Sum of all observations in the k groups 

= Sum of n₁ observations of first group + Sum of n₂ observations of second group + ... + Sum of n_k observations of kth group

= $n_1\overline{X_1}+n_2\overline{X_2}+...+n_k\overline{X_k}$

= $\sum _{i=1}^k{n}_i\overline{X_i}$                      .....(1)

Total number of observations in the k groups  = $n_1+n_2+...+n_k=\sum _{i=1}^k{n}_i$           .....(2)

∴ Mean $\overline{X}$ of all the groups = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{all} \mathrm{observations} \mathrm{in} \mathrm{the} k \mathrm{groups}}{\mathrm{Total} \mathrm{number} \mathrm{of} \mathrm{observations} \mathrm{in} \mathrm{the} k \mathrm{groups}}=\frac{{\displaystyle \sum _{i=1}^k}{n}_i\overline{X_i}}{{\displaystyle \sum _{i=1}^k}{n}_i}$            [Using (1) and (2)]        

Hence, the correct answer is option (c).

Question 19:

The median of the data : 4,4,5,7,6,7,7,12,3 is 

(a) 4
(b) 5
(c) 6
(d) 7

Answer 19:

Arranging the data in ascending order, we have

3, 4, 4, 5, 6, 7, 7, 7, 12

Here, the number of observations n = 9, which is odd

∴ Median = Value of $\left(\frac{9+1}{2}\right)\mathrm{th}$ observation = Value of $\left(\frac{10}{2}\right)\mathrm{th}$ observation = Value of 5th observation = 6

Thus, the median of given data is 6.

Hence, the correct answer is option (c).

Question 20:

The median of the data : 78,56,22,34,45,54,39,68,54,84 is 

(a) 45
(b) 49.5
(c) 53.5
(d) 56

Answer 20:

Disclaimer: Options (d) in the question changed to match the answer.

The median of the data : 78,56,22,34,45,54,39,68,54,84 is 

(a) 45
(b) 49.5
(c) 53.5
(d) 54

Solution:

Arranging the data in ascending order, we have

22, 34, 39, 45, 54, 54, 56, 68, 78, 84

Here, the number of observations n = 10, which is even

∴ Median of the data

= $\frac{\mathrm{Value} \mathrm{of} \left({\displaystyle \frac{10}{2}}\right)\mathrm{th} \mathrm{observation}+\mathrm{Value} \mathrm{of} \left({\displaystyle \frac{10}{2}}+1\right)\mathrm{th} \mathrm{observation}}{2}$

= $\frac{\mathrm{Value} \mathrm{of} 5\mathrm{th} \mathrm{observation}+\mathrm{Value} \mathrm{of} 6\mathrm{th} \mathrm{observation}}{2}$

= $\frac{54+54}{2}$

= $\frac{108}{2}$

= 54

Thus, the median of the given data is 54.

Hence, the correct answer is option (d).

Question 21:

There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be -3.5. The mean of the given numbers is 

(a) 46.5
(b) 49.5
(c) 53.5
(d) 56.5

Answer 21:

Let the 50 numbers be x₁, x₂, x₃,..., x₅₀.

When each number is subtracted from 53, the resulting numbers are 53 − x₁, 53 − x₂, 53 − x₃,..., 53 − x₅₀.

It is given that, the mean of resulting numbers 53 − x₁, 53 − x₂, 53 − x₃,..., 53 − x₅₀ is −3.5.

We know

Mean = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

$\therefore -3.5=\frac{\left(53-x_1\right)+\left(53-x_2\right)+\left(53-x_3\right)+...+\left(53-x_{50}\right)}{50}$

$\Rightarrow \frac{\underset{50 \mathrm{times}}{\underbrace{53+53+53+...+53}}-\left(x_1+x_2+x_3+...+x_{50}\right)}{50}=-3.5$

$\Rightarrow \frac{53\times 50}{50}-\frac{x_1+x_2+x_3+...+x_{50}}{50}=-3.5$

$\Rightarrow \frac{x_1+x_2+x_3+...+x_{50}}{50}=53+3.5=56.5$        .....(1)

Now,

Mean of 50 given numbers = $\frac{x_1+x_2+x_3+...+x_{50}}{50}$

∴ Mean of 50 given numbers = 56.5               [From (1)]

Thus, the mean of the given numbers is 56.5.

Hence, the correct answer is option (d).

Question 22:

The mean of 100 observations is 50. If one of the observations which was 50 is replaced by 150, the resulting will be 

(a) 50.5
(b) 51
(c) 51.5
(d) 52

Answer 22:

We know

Mean = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

⇒ Sum of observations = Mean × Number of observations

Mean of 100 observations = 50      (Given)

∴ Sum of 100 observations = 50 × 100 = 5000

If one of the observations which was 50 is replaced by 150, then

New sum of 100 observations = Sum of 100 observations − 50 + 150

⇒ New sum of 100 observations = 5000 − 50 + 150 = 5100

∴ New mean of 100 observations = $\frac{\mathrm{New} \mathrm{sum} \mathrm{of} 100 \mathrm{observations}}{100}=\frac{5100}{100}=51$

Thus, the new mean of 100 observations is 51.

Hence, the correct answer is option (b).

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Question 23:

Mode of the data : 15,14,19,20,14,15,16,14,15,18,14,19,15,17,15 is 

(a) 14
(b) 15
(c) 16
(d) 17

Answer 23:

Arranging the given data in ascending order, we have

14, 14, 14, 14, 15, 15, 15, 15, 15, 16, 17, 18, 19, 19, 20

Mode is the value of observation which occurs most frequently in a set of observations.

It can be seen that the value 15 occurs maximum number of times i.e. 5.

Thus, the mode of given data is 15.

Hence, the correct answer is option (b).

Question 24:

The empirical relation between mean, mode and median is

(a) Mode = 3 Median − 2 Mean

(b) Mode = 2 Median − 3 Mean

(c) Median = 3 Mode − 2 Mean

(d) Mean = 3 Median − 2 Mode

Answer 24:

The relation between mean, median and mode is

Hence, the correct option is (a).

Question 25:

The mean of a, b, c, d and e is 28. If the mean of a, c, and e is 24, What is the mean of b and d?

(a) 31

(b) 32

(c) 33

(d) 34

Answer 25:

Given that the mean of a, b, c, d and e is 28. They are 5 in numbers.

Hence, we have

But, it is given that the mean of a, c and e is 24. Hence, we have

Then, we have

Hence, the mean of b and d is.

Hence, the correct choice is (d).

Question 26:

The mean of 25 observations is 36. Out of these observations if the mean of first 13 observations is 32 and that the last 13 observations is 40, the 13^th observations is 

(a) 23
(b) 36
(c) 38
(d) 40

Answer 26:

We know

Mean = $\frac{\mathrm{Sum} \mathrm{of} \mathrm{observations}}{\mathrm{Number} \mathrm{of} \mathrm{observations}}$

⇒ Sum of observations = Mean × Number of observations

Mean of 25 observations = 36                  (Given)

∴ Sum of 25 observations = 36 × 25 = 900

Mean of first 13 observations = 32           (Given)

∴ Sum of first 13 observations = 32 × 13 = 416

Mean of last 13 observations = 40           (Given)

∴ Sum of last 13 observations = 40 × 13 = 520

Now,

13^th observation

= Sum of first 13 observations + Sum of last 13 observations − Sum of 25 observations

= 416 + 520 − 900

= 36

Thus, the 13^th observation is 36.

Hence, the correct answer is option (b).